Kirchhoff’s first rule – problems and solutions

1. Based on figure as shown below, if I₁ = 5A, I₂ = 3 A, I₃ = 6 A, then I

₄ = ………

Solution :

Kirchhoff’s first rule states that the sums of all currents entering the junction must equal the sum of all currents leaving the junction.

I₁ and I₂ are entering the junction, whereas I₃ and I₄ are leaving.

Apply Kirchhoff’s first rule :

I₁ + I₂ = I₃ + I₄

5 + 3 = 6 + I₄

8 = 6 + I₄

I₄ = 8 – 6

I₄ = 2 Ampere

2. If I₁ = I₃ then I₄ =……..

Known :

I₁ = I₃ = 2 Ampere

I₂ = 3 Ampere

I₅ = 4 Ampere

Wanted: I₄

Solution :

I₁ and I₄ are entering the junction, whereas I₂, I₃, and I₅ are leaving.

Apply Kirchhoff’s first rule :

I₁ + I₄ = I₂ + I₃ + I₅

2 + I₄ = 3 + 2 + 4

2 + I₄ = 9

I₄ = 9 – 2

I₄ = 7 Ampere

3. I₁ = …….

Solution :

I am entering the junction, whereas I₁, I₂, and I₃ are leaving.

Apply Kirchhoff’s first rule :

I = I₁ + I₂ + I₃

7 A = I₁ + 1 A + 4 A

7 A = I₁ + 5 A

7 A – 5 A = I₁

I₁ = 2 Ampere

4. I₂ = ………

Wanted: I₂

Solution :

I₁ is entering the junction, whereas I_2, I_3, and I₄ are leaving.

Apply Kirchhoff’s first rule :

I₁ = I₂ + I₃ + I₄

6 = I₂ + 3 + 1

6 = I₂ + 4

I₂ = 6 – 4

I₂ = 2 Ampere

5. If I₂ = 1/4 I₁ , then I₃ = ………

Known :

I₁ = 600 milliAmpere

Wanted: I₃

Solution :

I₂ = 1/4 I₁ = 1/4 (600 mA) = 150 mA

I₁ is entering the junction, whereas I₂ and I₃ are leaving.

Apply Kirchhoff’s first rule :

I₁ = I₂ + I₃

600 mA = 150 mA + I₃

I₃ = 600 mA – 150 mA

I₃ = 450 mA

6. Based on figure below, find I!

Solution :

Apply Kirchhoff‘s first rule :

3 + 5 + 2 = 7 + I

10 = 7 + I

10 – 7 = I

I = 3 Ampere

7. Based on figure below, if I₁ = 4 A, I₂ = 3 A, I₄ = 7 A, I₅ = 4 A, determine I₃.

Solution :

Electric current entering the junction = I₁ + I₃ = 4 A + I₃

Electric current leaving the junction = I₂ + I₄ + I₅ = 3 A + 7 A + 4 A = 14 A

14 A = 4 A + I₃

I₃ = 14 A – 4 A

I₃ = 10 A